Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{x^2 + 8x - 20}{-5x - 25} \times \dfrac{3x + 15}{2x + 20} $
First factor the quadratic. $z = \dfrac{(x + 10)(x - 2)}{-5x - 25} \times \dfrac{3x + 15}{2x + 20} $ Then factor out any other terms. $z = \dfrac{(x + 10)(x - 2)}{-5(x + 5)} \times \dfrac{3(x + 5)}{2(x + 10)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (x + 10)(x - 2) \times 3(x + 5) } { -5(x + 5) \times 2(x + 10) } $ $z = \dfrac{ 3(x + 10)(x - 2)(x + 5)}{ -10(x + 5)(x + 10)} $ Notice that $(x + 5)$ and $(x + 10)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 3\cancel{(x + 10)}(x - 2)(x + 5)}{ -10(x + 5)\cancel{(x + 10)}} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $z = \dfrac{ 3\cancel{(x + 10)}(x - 2)\cancel{(x + 5)}}{ -10\cancel{(x + 5)}\cancel{(x + 10)}} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $z = \dfrac{3(x - 2)}{-10} $ $z = \dfrac{-3(x - 2)}{10} ; \space x \neq -10 ; \space x \neq -5 $